Question: Multiply the following complex numbers: $({5i}) \cdot ({3-4i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({5i}) \cdot ({3-4i}) = $ $ ({0} \cdot {3}) + ({0} \cdot {-4}i) + ({5}i \cdot {3}) + ({5}i \cdot {-4}i) $ Then simplify the terms: $ (0) + (0i) + (15i) + (-20 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 0 + (0 + 15)i - 20i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 0 + (0 + 15)i - (-20) $ The result is simplified: $ (0 + 20) + (15i) = 20+15i $